When $ k \ge 10$, $k!$ must be a multiple of $100$ because its prime factorization contains two $5$s and more than two $2$s. Therefore, the desired result equals the last two digits of $$1! + 2! + \cdots + 9! \pmod{100}$$

Let's compute these terms separately:

$$\begin{array}{lcrl} 1!&\equiv& 1&\pmod{100} \\ 2!&\equiv& 2 & \pmod{100} \\ 3!&\equiv& 6 &\pmod{100} \\ 4!&\equiv& 24 &\pmod{100} \\ 5!&\equiv& 20 & \pmod{100} \\ 6!&\equiv& 20 & \pmod{100} \\ 7!&\equiv&40 & \pmod{100} \\ 8!&\equiv & 20 & \pmod{100}\\9!&\equiv&80 & \pmod{100} \end{array}$$

Adding the numbers on the right side leads to the result $\boxed{13}$.